Answer 1

When three coins are tossed once, the sample space is given by
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
i.e. n (S) = 8

 Let E₁ = event of getting exactly two heads
Then E₁ = {HHT, HTH, THH}
i.e. n(E₁) = 3

\[\therefore P\left( E_1 \right) = \frac{n\left( E_1 \right)}{n\left( S \right)} = \frac{3}{8}\]